Find the number of ways of dividing 6 couples in 3 groups if each group has exactly one couple and each group has 2 males and 2 females.

Explanation

There are 6 couples `(A_(1)B_(1)),(A_(2)B_(2)),(A_(3)B_(3)),(A_(4)B_(4)),(A_(5)B_(5)) " and " (A_(6)B_(6)), " where " A_(i)` is male and `B_(i)` is female.
We have to divide these couples (12 persons) in three groups P, Q and R, such that each containing exactly one couple having two males and two females.
First select three couples for three groups P, Q, R. Number of ways are `""^(6)C_(3)`.
Now from the remaining three couples three males can enter in groups P, Q and R in 3! ways.
The remaining three females cannot enter the group in which their spouses have entered.
So, for females we have derangement
`D(3)=3!(1-(1)/(1!)+(1)/(2!)-(1)/(3!))=2`
Therefore, total number of required ways `= ""^(6)C_(3)xx3!xx2`
=240